Question

A sample of pure NO2NO2 is heated to 335 ∘C∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g)2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.525 g/Lg/L at 0.750 atmatm .Calculate Kc for the reaction

Answer 1

Answer:

0.97.

Explanation:

So, let us first write out the balanced equilibrium equation below;

2 NO2(g) <============> 2 NO(g) + O2(g).

So, from the above equation of reaction 2 moles of NO2 gives us 3 moles of NO and one mole of O2.

Let us say that initially the concentration NO2 = 1 and that that NO and O2 is zero respectively. Then, at specific time, t = t the concentration of NO2 = 1 - 2x and the concentration of NO and O2 = 2x and x respectively.

So, kc = [2x]^2 [x] / [ 1 - 2x]^2.

Now, we have two unknown variables that is the value of x and kc. So, we will have to find the value of x first before we can proceed to solve for kc.

Recall the ideal gas law equation which is; PV = nRT. Where P= pressure, V= volume, R= gas constant, n= number of moles and T = temperature.

Also, n= mass/ molar mass.

So, PV = (mass/ molar mass) × R × T.

We are given the density of the gas mixture to be = 0.525 g/L. Where density = mass/ volume. And pressure = 0.750 atm.

Hence, P =( density × R × T)/ Molar mass.

Molar mass = 0.525 × 0.0821 × (335° C + 273)/ 0.750.

Molar mass= 34.9 g/ mol.

The next thing to do is to find the value of x.

Therefore;

Molar mass of the mixture= Molar mass of NO2 × (1 -2x / 1 + x) + molar mass of NO ( 2x/ 1 + x) + molar mass of O2 ( x/ 1 + x).

Therefore, solving for x gives;

34.9 = 46/ 1 + x.

34.9 + 34.9 x = 46.

34.9 x = 46 - 34.9.

x= 0.318.

So, kc = (0.6361)^2 (0.318) / (0.3639)^2.

Kc= 0.4046 × 0.318/ 0.1324.

kc = 0.97.

Answer 2

Answer:

$k_c =$ 1. 1 × 10⁻²

Explanation:

Given that:

Temperature = 335 ° C = (335+ 273)K = 608

Pressure = 0.750 atm

Volume = 1 Litre

number of moles of NO2 = ???

Rate Constant =0.0821 L atm /K/mol

Using the Ideal gas equation

PV = nRT

n = $\frac{PV}{RT}$

n = $\frac{0.75*1}{0.0821*608}$

n = 0.015

n = 1.5 × 10⁻² mole

Density = 0.525 g/L

The equation for the reaction can be illustrated as:

                     2NO2(g)         ⇌          2NO(g)         +         O2(g)

For the ICE table; we have:

 

Initial                 x                                   0                              0

Change            -2y                               + 2y                          +y

Equilibrium        (x - 2y)                        2y                             y

Total moles at equilibrium = (x-2y)+2y+y

= x + y moles

However,

1.5 × 10⁻² mole of the mixture has a mass of 0.525 g

i.e x + y moles = 1.5 × 10⁻² mole

Now, molar mass of 1 mole of NO2 = 46g/mol

Since number of moles = $\frac{mass}{molar mass}$

mass of (x-2y) moles = 46 × (x-2y) g

Molar mass of NO = 30 g/mol

Also, mass of NO = 2y × 30 = 60y

Molar mass of O2 = 32 g/mol

Mass of O2 = y × 32 = 32y

Total mass = ( 46x - 90y)+60y+32y = 0.525

46x = 0.525

x = $\frac{0.525}{46}$

x = 0.0114

x = 1.14 × 10⁻²

x + y moles = 1.5 × 10⁻²

y =  1.5 × 10⁻² -  1.14 × 10⁻²

y = 0.0036

y = 3.6 × 10⁻³

At equilibrium

[NO2] = ( 1.14 - 2(0.36))× 10⁻² = 4.2 × 10⁻³ M

[NO] = 2 ( 3.6 × 10⁻³)  = 7.2 × 10⁻³ M

[O2] = 3.6 × 10⁻³ M

$k_c = \frac{[NO]^2[O_2]}{[NO_2]^2}$

$k_c = \frac{(7.2*10^{-3})^2(3.6*10^{-3})}{(4.2*10^-3)^2}$

$k_c =$ 0.011

$k_c =$ 1. 1 × 10⁻²